If A B and tan A - tan B = x and cot B - co | vedclass

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(C) Given,$	an A - 	an B = x$ $\cot B - \cot A = y$ $\Rightarrow \frac{1}{	an B} - \frac{1}{	an A} = y$ $\Rightarrow \frac{	an A - 	an B}{	an A

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$\frac{1}{x} + \frac{1}{y}$

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(C) Given,$	an A - 	an B = x$ $\cot B - \cot A = y$ $\Rightarrow \frac{1}{	an B} - \frac{1}{	an A} = y$ $\Rightarrow \frac{	an A - 	an B}{	an A 	an B} = y$ $\Rightarrow \frac{x}{	an A 	an B} = y$ $\Rightarrow 	an A 	an B = \frac{x}{y}$ Now,$\cot (A - B) = \frac{1}{	an (A - B)} = \frac{1 + 	an A 	an B}{	an A - 	an B}$ Substituting the values,we get: $\cot (A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{\frac{y + x}{y}}{x} = \frac{x + y}{xy} = \frac{x}{xy} + \frac{y}{xy} = \frac{1}{y} + \frac{1}{x}$

If $A > B$ and $\tan A - \tan B = x$ and $\cot B - \cot A = y$,then $\cot (A - B) = $

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